# Homework Help: Frictionless pully acceleration

1. Mar 4, 2013

### CollegeStudent

1. The problem statement, all variables and given/known data
M1 is on a frictionless flat horizontal surface. It is attached to M2 by a string, which hangs freely from the string which passes over a frictionless pulley of negligible mass. Find a general expression for the acceleration of both M1 and M2. Look at your result and argue that it should be correct.

2. Relevant equations

3. The attempt at a solution

I'm not really sure how to word this but:

If the mass of M2 is...say .050kg , the force acting on M2 would be the mass of it * gravity. In order to accelerate, M1 would need to have a force act on it that is greater than the mass of M2. If the force pushing M1 was = to the mass of M2 , then M1 would be static.

Is this correct?

2. Mar 4, 2013

### haruspex

A force cannot be compared (as in greater, less than) to a mass.
Let the tension in the string be T. List the forces acting on each mass and write out the appropriate ƩF=ma equations for each.

3. Mar 4, 2013

### CollegeStudent

Okay...so for the hanging M2 we have

ƩF=ma
ƩF=m * g

And M1 has some type of force acting on it that keeps it from rolling back towards M2. (Gravity and normal force are excluded here because there is no vertical acceleration of M1)

That force would be the opposite of the force acting on M2.

That means that the tension of the string connecting the 2...would equal the force pushing M1 foward.

So if tension = the force on M1...and the force of M1 is equal and opposite to the force acting on M2

ƩF = T = mass of M2 * g

If that's true...that would make M1 be static and not moving foward nor backwards...so in order for M1 to accelerate foward...the magnitude of the force acting on M1 would have to be GREATER than the mass of M2 * g.

Does THIS look any better?

4. Mar 4, 2013

### haruspex

No, mg is a force, so it's one part of the ƩF. Let's say the mass of M2 is m2, so the gravitational force here is m2g. What else is in ƩF?
Suppose the system accelerates at rate a; so M2 accelerates down at rate a (while M1 accelerates towards the pulley at the same rate). So the RHS is just m2a.
What makes you think that? It will accelerate towards the pulley - there's nothing holding it back.
No, you must not assume T = m2g. If that were true, there'd be no net force on M2, so no acceleration.

5. Mar 5, 2013

### CollegeStudent

And that isn't the same thing as saying the acceleration is going to be the opposite of the mass of M2 times gravity?

Right, sorry I wasn't clear...I meant there has to be some force acting on M1 to prevent it from rolling back. Otherwise, yes it will accelerate towards the pulley.

Ahh okay...so when tension = the force acting on M2..the 2 objects are static? It just throws me off because...if you have a force reader, (those things you can attach to a string and pull and see the amount of force being applied) wouldn't the forces be equal and opposite? Thus making the tension = to that force?

Then again that makes the string static...okay, back to the original problem....so in order for M1 to accelerate either forwards or backwards the force acting one one side, either M1 or M2 would have to be greater than the force acting on the opposing object?

6. Mar 5, 2013

### MostlyHarmless

Draw two free body diagrams. The first "body" will be M1, the only force acting on M1 is tension towards the pully. Your second "body" is M2. There are two forces acting on M2, Tension pulling up, and the force due to gravity pulling down.

Now set up equations for both Free body diagrams using newtons 2nd law. ƩF=ma